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A pair of dice is rolled. Find the probability of rolling a) The probability of rolling a sum not more than 9 is (Type an int

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Answer #1

A pair of dice is rolled.

So

Sample space

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

n(S) = 36

1) Let A be the event of sum not more than 9

A = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5)

(5,1) (5,2) (5,3) (5,4)

(6,1) (6,2) (6,3) }

n(A) = 30

P(A) = n(A) /n(S) = 30/36 = 1/6

2)

Let B be the event of sum not more than 6

B = {(1,1) (1,2) (1,3) (1,4) (1,5)

(2,1) (2,2) (2,3) (2,4)

(3,1) (3,2) (3,3)

(4,1) (4,2)

(5,1)}

n(B) = 15

P(B) = n(B) /n(S) = 15/36 = 5/12

3)

Let C be the event of sum between 6 and 10

C = {(1,6)

(2,5) (2,6)

(3,4) (3,5) (3,6)

(4,3) (4,4) (4,5)

(5,2) (5,3) (5,4)

(6,1) (6,2) (6,3) }

n(C) = 15

P(C) = n(C) /n(S) = 15/36 = 5/12

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Answer #2

SOLUTION :


Pair of dice rolled.


Only part(a) is required to be answered.


P(sum is not greater than 9)

= P( sum ≤ 9) 

= 1 - P(sum > 9)


Total space = 6 * 6 = 36 events


Events,  (sum >  9 ) are :


(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6,6) = 6 events


So, P( sum > 9) = 6/36 = 1/6


Hence, P (sum ≤ 9) = 1 - 1/6 = 5/6 (ANSWER).

answered by: Tulsiram Garg
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